molar mass of C4H10 = 58.1 g/mol
molar mass of O2 = 32 g/mol
number of mol = (given mass)/(molar mass)
number of mol of C4H10 = 8.16/58.1
= 0.141 mol
number of mol of O2 = 15.7/32
= 0.491 mol

reaction taking place
2C4H10 + 13O2 --> 8CO2 + 10H2O
according to reaction
13 mol of O2 required 2 mol of C4H10
1 mol of O2 required 2/13 mol of C4H10
0.491 mol of O2 required (2/13)*0.491 mol of C4H10
0.491 mol of O2 required 0.0755 mol of C4H10

but we have only 0.141 mol of C4H10
so, C4H10 is in excess

again,
13 mol of O2 give 8 mol of CO2
1 mol of O2 give 8/13 mol of CO2
0.491 mol of O2 give (8/13)*0.491 mol of CO2
0.491 mol of O2 give 0.302 mol of CO2
so,
number of mol of CO2 formed = 0.302 mol

moalar mass of CO2 = 44.0 g/mol
mass of CO2 = (number of mol of CO2 formed)*(moalar mass of CO2)
= 0.302*44.0
= 13.3 g

formula for limiting reagent is O2

number of mol of C4H10 remains = (number of mol of C4H10) - (number of mol of C4H10 reacted)
= (0.141 - 0.0755)
= 0.0655 mol

mass of C4H10remains = (number of mol of C4H10 remains)*(molar mass of C4H10)
= 0.0655*58.1
= 3.81 g